Question

So sánh:

a) \(\int\limits^1_0\left(2x+3\right)dx\) và \(\int\limits^1_02xdx+\int\limits^1_03dx\);

b) \(\int\limits^1_0\left(2x-3\right)dx\) và \(\int\limits^1_02xdx-\int\limits^1_03dx\).

Answer 1

a) \(\int\limits_0^1 {(2x + 3)dx} = \left. {\left( {{x^2} + 3x} \right)} \right|_0^1 = 1 + 3 = 4\).

\(\int\limits_0^1 {2xdx} + \int\limits_0^1 {3dx} = \left. {{x^2}} \right|_0^1 + \left. {3x} \right|_0^1 = 1 + 3 = 4\).

Vậy \(\int\limits_0^1 {(2x + 3)dx} \) = \(\int\limits_0^1 {2xdx} + \int\limits_0^1 {3dx} \).

b) \(\int\limits_0^1 {(2x - 3)dx} = \left. {\left( {{x^2} - 3x} \right)} \right|_0^1 = 1 - 3 = - 2\).

\(\int\limits_0^1 {2xdx} - \int\limits_0^1 {3dx} = \left. {{x^2}} \right|_0^1 - \left. {3x} \right|_0^1 = 1 - 3 = - 2\).

Vậy \(\int\limits_0^1 {(2x - 3)dx} \) = \(\int\limits_0^1 {2xdx} - \int\limits_0^1 {3dx} \).